Index of submodule generated by multiplication by one element of free $\mathbb{Z}$-module

Question

This question appears in an algebraic number theory notes with $M=\mathcal{O}_K$ for a number field $K$.

Let $M$ be a free $\mathbb{Z}$-module of rank $n$ which is a ring itself. Let $x\in M$, then multiplication by $x$, $\phi_x:y\mapsto xy$ is a $M$-endomorphism, hence can be represented by a $n\times n$ $\mathbb{Z}$-matrix C. Let $\{v_1\,\dots,v_n\}$ be a basis of $M$, and let $N$ be the submodule generated by $\{xv_1,\dots,xv_n\}$ over $\mathbb{Z}$. I want to show that $[M:N]=|\det(C)|$, where $[M:N]$ is the index of $N$ as a subgroup of $M$. Could anyone please help?

Answer 1

Note that $N$ is just the image of $\phi_x$.

Let more generally $M$ be free abelian group of finite rang and let $\phi$ be an injective endomorphism of $M$. Then $$ \lvert M / {\operatorname{im}(\phi)} \rvert = \lvert \det(\phi) \rvert. $$ To show this, we may assume that $M = \mathbb{Z}^n$. The endomorphism $\phi$ is then given by multiplication with some $\mathbb{Z}$-matrix $C$ of size $n \times n$. The matrix $C$ has a Smith normal form. There hence exist invertible $\mathbb{Z}$-matrices $S$ and $T$ of size $n \times n$ such that $$ C = S D T $$ for some diagonal matrix $$ D = \begin{pmatrix} d_1 & & \\ & \ddots & \\ & & d_n \end{pmatrix}, $$ with $d_1, \dotsc, d_n > 0$. (We may also assume that $d_i$ divides $d_{i+1}$ for all $i = 1, \dotsc, n-1$, but we won’t need this.) Both $S$ and $T$ have determinant $\pm 1$ because they are invertible integer matrices. It follows that \begin{align*} \lvert \det(\phi) \rvert &= \lvert \det(C) \rvert \\ &= \lvert \det(SDT) \rvert \\ &= \lvert \det(S) \det(D) \det(T) \rvert \\ &= \lvert \det(S) \rvert \cdot \lvert \det(D) \rvert \cdot \lvert \det(T) \rvert \\ &= d_1 \dotsm d_n . \end{align*} We find on the other hand that $$ M / \operatorname{im}(\phi) = M / C M = M / SDT M = M / SD M $$ because $T$ is invertible and thus $TM = M$. The isomorphism of abelian groups $$ M \to M, \quad x \mapsto Sx $$ maps $DM$ onto $SDM$, and thus descends to an isomorphism of abelian groups between $M / DM$ and $M / SD M$. We hence find that \begin{align*} M / SD M &\cong M / DM \\ &= \mathbb{Z}^n / D \mathbb{Z}^n \\ &= ( \mathbb{Z} \oplus \dotsb \oplus \mathbb{Z} ) / (d_1 \mathbb{Z} \oplus \dotsb \oplus d_n \mathbb{Z}) \\ &\cong \mathbb{Z} / d_1 \mathbb{Z} \oplus \dotsb \mathbb{Z} / d_n \mathbb{Z}. \end{align*} This shows overall that $$ \lvert M / \operatorname{im}(\phi) \rvert = \lvert \mathbb{Z} / d_1 \mathbb{Z} \oplus \dotsb \oplus \mathbb{Z} / d_n \mathbb{Z} \rvert = d_1 \dotsm d_n. $$

Answer 2

From the isomorphism $M\to \Bbb{Z}^n$ given by the $v_j$ then $M/N\cong \Bbb{Z}^n/ C \Bbb{Z}^n$ has $Vol( C[0,1]^n) = |\det(C)|$ elements.